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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.Since $h$ and $m$ are constants,$\lambda \propt

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Decrease by $\lambda/2$

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.Since $h$ and $m$ are constants,$\lambda \propto \frac{1}{v}$.Let the initial velocity be $v_1 = v$ and the initial wavelength be $\lambda_1 = \lambda$.When the velocity is doubled,the new velocity is $v_2 = 2v$.The new wavelength is $\lambda_2 = \frac{h}{m(2v)} = \frac{1}{2} \left( \frac{h}{mv} \right) = \frac{\lambda}{2}$.The change in wavelength is $\Delta \lambda = \lambda_1 - \lambda_2 = \lambda - \frac{\lambda}{2} = \frac{\lambda}{2}$.Thus,the wavelength decreases by $\frac{\lambda}{2}$.

If the velocity of a free electron is doubled,the change in its de Broglie wavelength will be ...

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